/* 1050:To the Max */
/*
描述
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:

9 2
-4 1
-1 8
and has a sum of 15.
输入
The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 500. The numbers in the array will be in the range [-127,127].
输出
Output the sum of the maximal sub-rectangle.
*/


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using std::max;

const int MAXN = 100 + 1;
const int INF = 0x3f3f3f3f;

int N;
int nMatrix[MAXN][MAXN];

void Read()
{
    for (int i = 1; i <= N; ++i)
    {
        for (int j = 1; j <= N; ++j)
        {
            scanf("%d", &nMatrix[i][j]);
        }
    }
}

void Dp()
{
    int nRet = -INF;
    int nBuf[MAXN];

    for (int i = 1; i <= N; ++i)
    {
        memset(nBuf, 0, sizeof(nBuf));
        for (int j = i; j <= N; ++j)
        {
            for (int k = 1; k <= N; ++k)
            {
                nBuf[k] += nMatrix[j][k];
            }

            int dp = 0;
            for (int k = 1; k <= N; ++k)
            {
                dp = max(nBuf[k], dp + nBuf[k]);
                nRet = max(nRet, dp);
            }
        }
    }

    printf("%d\n", nRet);
}

int main()
{
    while (scanf("%d", &N) == 1)
    {
        Read();
        Dp();
    }

    return 0;
}

// 结果正确，历经磨难依旧还没有通过。。。原因待查
